The costs involved with a health insurance plan can be confusing so I perform an analysis of different options to find which plan is most cost effective
My wife and I recently brought a new R programmer into our family so we had to update our health insurance. Becky is a researcher in neuroscience and psychology at NYU so we decided to choose an NYU insurance plan.
For families there are two main plans: Value and Advantage. The primary differences between the plans are the following:
Value Plan Amount
Advantage Plan Amount
The amount we pay every other week in order to have insurance
$160 ($4,160 annually)
$240 ($6,240 annually)
Amount we pay directly to health providers before the insurance starts covering costs
After the deductible is met, we pay this percentage of medical bills
This is the most we will have to pay to health providers in a year (premiums do not count toward this max)
We put them into a tibble for use later.
# use tribble() to make a quick and dirty tibble
parameters <- tibble::tribble(
~Plan, ~Premiums, ~Deductible, ~Coinsurance, ~OOP_Maximum,
'Value', 160*26, 1000, 0.2, 6000,
'Advantage', 240*26, 800, 0.1, 5000
Other than these cost differences, there is not any particular benefit of either plan over the other. That means whichever plan is cheaper is the best to choose.
This blog post walks through the steps of evaluating the plans to figure out which to select. Code is included so anyone can repeat, and improve on, the analysis for their given situation.
In order to figure out which plan to select we need to figure out the all-in cost, which is a function of how much we spend on healthcare in a year (we have to estimate our annual spending) and the aforementioned premiums, deductible, coinsurance and out-of-pocket maximum.
#' @title cost
#' @description Given healthcare spend and other parameters, calculate the actual cost to the user
#' @details Uses the formula above to caluclate total costs given a certain level of spending. This is the premiums plus either the out-of-pocket maximum, the actual spend level if the deductible has not been met, or the amount of the deductible plus the coinsurance for spend above the deductible but below the out-of-pocket maximum.
#' @author Jared P. Lander
#' @param spend A given amount of healthcare spending as a vector for multiple amounts
#' @param premiums The annual premiums for a given plan
#' @param deductible The deductible for a given plan
#' @param coinsurance The coinsurance percentage for spend beyond the deductible but below the out-of-pocket maximum
#' @param oop_maximum The maximum amount of money (not including premiums) that the insured will pay under a given plan
#' @return The total cost to the insured
#' cost(3000, 4160, 1000, .20, 6000)
#' cost(3000, 6240, 800, .10, 5000)
cost <- function(spend, premiums, deductible, coinsurance, oop_maximum)
# spend is vectorized so we use pmin to get the min between oop_maximum and (deductible + coinsurance*(spend - deductible)) for each value of spend provided
# we can never pay more than oop_maximum so that is one side
# if we are under oop_maximum for a given amount of spend,
# this is the cost
deductible + coinsurance*(spend - deductible)
# we add the premiums since that factors into our cost
With this function we can see if one plan is always, or mostly, cheaper than the other plan and that’s the one we would choose.
For the rest of the code we need these R packages.
We call our cost function on each amount of spend for the Value and Advantage plans.
spending <- spending %>%
# use our function to calcuate the cost for the value plan
# use our function to calcuate the cost for the Advantage plan
# compute the difference in costs for each plan
# the winner for a given amount of spend is the cheaper plan
mutate(Winner=if_else(Advantage < Value, 'Advantage', 'Value'))
The results are in the following table, showing every other row to save space. The Spend column is a theoretical amount of spending with a red bar giving a visual sense for the increasing amounts. The Value and Advantage columns are the corresponding overall costs of the plans for the given amount of Spend. The Difference column is the result of Advantage – Value where positive numbers in blue mean that the Value plan is cheaper while negative numbers in red mean that the Advantage plan is cheaper. This is further indicated in the Winner column which has the corresponding colors.
Of course, plotting often makes it easier to see what is happening.
select(Spend, Value, Advantage) %>%
# put the plot in longer format so ggplot can set the colors
gather(key=Plan, value=Cost, -Spend) %>%
ggplot(aes(x=Spend, y=Cost, color=Plan)) +
scale_color_brewer(type='qual', palette='Set1') +
labs(x='Healthcare Spending', y='Out-of-Pocket Costs') +
It looks like there is only a small window where the Advantage plan is cheaper than the Value plan. This will be more obvious if we draw a plot of the difference in cost.
ggplot(aes(x=Spend, y=Difference, color=Winner, group=1)) +
geom_hline(yintercept=0, linetype=2, color='grey50') +
y='Difference in Out-of-Pocket Costs Between the Two Plans'
scale_color_brewer(type='qual', palette='Set1') +
To calculate the exact cutoff points where one plan becomes cheaper than the other plan we have to solve for where the two curves intersect. Due to the out-of-pocket maximums the curves are non-linear so we need to consider four cases.
The spending exceeds the point of maximum out-of-pocket spend for both plans
The spending does not exceed the point of maximum out-of-pocket spend for either plan
The spending exceeds the point of maximum out-of-pocket spend for the Value plan but not the Advantage plan
The spending exceeds the point of maximum out-of-pocket spend for the Advantage plan but not the Value plan
When the spending exceeds the point of maximum out-of-pocket spend for both plans the curves are parallel so there will be no cross over point.
When the spending does not exceed the point of maximum out-of-pocket spend for either plan we set the cost calculations (not including the out-of-pocket maximum) for each plan equal to each other and solve for the amount of spend that creates the equality.
To keep the equations smaller we use variables such as \(d_v\) for the Value plan deductible, \(c_a\) for the Advantage plan coinsurance and \(oop_v\) for the out-of-pocket maximum for the Value plan.
When the spending exceeds the point of maximum out-of-pocket spend for the Value plan but not the Advantage plan, we set the out-of-pocket maximum plus premiums for the Value plan equal to the cost calculation of the Advantage plan.
#' @title calculate_crossover_points
#' @description Given healthcare parameters for two plans, calculate when one plan becomes more expensive than the other.
#' @details Calculates the potential crossover points for different scenarios and returns the ones that are true crossovers.
#' @author Jared P. Lander
#' @param premiums_1 The annual premiums for plan 1
#' @param deductible_1 The deductible plan 1
#' @param coinsurance_1 The coinsurance percentage for spend beyond the deductible for plan 1
#' @param oop_maximum_1 The maximum amount of money (not including premiums) that the insured will pay under plan 1
#' @param premiums_2 The annual premiums for plan 2
#' @param deductible_2 The deductible plan 2
#' @param coinsurance_2 The coinsurance percentage for spend beyond the deductible for plan 2
#' @param oop_maximum_2 The maximum amount of money (not including premiums) that the insured will pay under plan 2
#' @return The amount of spend at which point one plan becomes more expensive than the other
#' 160, 1000, 0.2, 6000,
#' 240, 800, 0.1, 5000
calculate_crossover_points <- function(
premiums_1, deductible_1, coinsurance_1, oop_maximum_1,
premiums_2, deductible_2, coinsurance_2, oop_maximum_2
# calculate the crossover before either has maxed out
neither_maxed_out <- (premiums_2 - premiums_1 +
deductible_2*(1 - coinsurance_2) -
deductible_1*(1 - coinsurance_1)) /
(coinsurance_1 - coinsurance_2)
# calculate the crossover when one plan has maxed out but the other has not
one_maxed_out <- (oop_maximum_1 +
premiums_1 - premiums_2 +
# calculate the crossover for the reverse
other_maxed_out <- (oop_maximum_2 +
premiums_2 - premiums_1 +
# these are all possible points where the curves cross
all_roots <- c(neither_maxed_out, one_maxed_out, other_maxed_out)
# now calculate the difference between the two plans to ensure that these are true crossover points
all_differences <- cost(all_roots, premiums_1, deductible_1, coinsurance_1, oop_maximum_1) -
cost(all_roots, premiums_2, deductible_2, coinsurance_2, oop_maximum_2)
# only when the difference between plans is 0 are the curves truly crossing
all_roots[all_differences == 0]
We then call the function with the parameters for both plans we are considering.
We see that the Advantage plan is only cheaper than the Value plan when spending between $20,000 and $32,000.
The next question is will our healthcare spending fall in that narrow band between $20,000 and $32,000 where the Advantage plan is the cheaper option?
Probability of Spending
This part gets tricky. I’d like to figure out the probability of spending between $20,000 and $32,000. Unfortunately, it is not easy to find healthcare spending data due to the opaque healthcare system. So I am going to make a number of assumptions. This will likely violate a few principles, but it is better than nothing.
Assumptions and calculations:
Healthcare spending follows a log-normal distribution
We will work with New York State data which is possibly different than New York City data
We know the mean for New York spending in 2014
We will use the accompanying annual growth rate to estimate mean spending in 2019
We have the national standard deviation for spending in 2009
In order to figure out the standard deviation for New York, we calculate how different the New York mean is from the national mean as a multiple, then multiply the national standard deviation by that number to approximate the New York standard deviation in 2009
We use the growth rate from before to estimate the New York standard deviation in 2019
We then take just New York spending for 2014 and multiply it by the corresponding growth rate.
ny_spend <- health_spend %>%
# get just New York
filter(State_Name == 'New York') %>%
# this row holds overall spending information
filter(Item == 'Personal Health Care ($)') %>%
# we only need a few columns
select(Y2014, Growth=Average_Annual_Percent_Growth) %>%
# we have to calculate the spending for 2019 by accounting for growth
# after converting it to a percentage
mutate(Y2019=Y2014*(1 + (Growth/100))^5)
We see that the New York average is 1.4187464 times the national average. So we multiply the national standard deviation from 2009 by this amount to estimate the New York State standard deviation and assume the same annual growth rate as the mean. Recall, we can multiply the standard deviation by a constant.
My original assumption was that spending would follow a normal distribution, but New York’s resident agricultural economist, JD Long, suggested that the spending distribution would have a floor at zero (a person cannot spend a negative amount) and a long right tail (there will be many people with lower levels of spending and a few people with very high levels of spending), so a log-normal distribution seems more appropriate.
So we only have a 2.35% probability of our spending falling in that band where the Advantage plan is more cost effective. Meaning we have a 97.65% probability that the Value plan will cost less over the course of a year.
We can also calculate the expected cost under each plan. We do this by first calculating the probability of spending each (thousand) dollar amount (since the log-normal is a continuous distribution this is an estimated probability). We multiply each of those probabilities against their corresponding dollar amounts. Since the distribution is log-normal we need to exponentiate the resulting number. The data are on the thousands scale, so we multiply by 1000 to put it back on the dollar scale. Mathematically it looks like this.
The following code calculates the expected cost for each plan.
# calculate the point-wise estimated probabilities of the healthcare spending
# based on a log-normal distribution with the appropriate mean and standard deviation
# compute the expected cost for each plan
# and the difference between them
# exponentiate the numbers so they are on the original scale
# the spending data is in increments of 1000
# so multiply by 1000 to get them on the dollar scale
mutate_each(funs=~ .x * 1000)
This shows that overall the Value plan is cheaper by about $1,324 dollars on average.
We see that there is a very small window of healthcare spending where the Advantage plan would be cheaper, and at most it would be about $600 cheaper than the Value plan. Further, the probability of falling in that small window of savings is just 2.35%.
So unless our spending will be between $20,000 and $32,000, which it likely will not be, it is a better idea to choose the Value plan.
Since the Value plan is so likely to be cheaper than the Advantage plan I wondered who would pick the Advantage plan. Economist Jon Hersh invokes behavioral economics to explain why people may select the Advantage plan. Some parts of the Advantage plan are lower than the Value plan, such as the deductible, coinsurance and out-of-pocket maximum. People see that under certain circumstances the Advantage plan would save them money and are enticed by that, not realizing how unlikely that would be. So they are hedging against a low probability situation. (A consideration I have not accounted for is family size. The number of members in a family can have a big impact on the overall spend and whether or not it falls into the narrow band where the Advantage plan is cheaper.)
In the end, the Value plan is very likely going to be cheaper than the Advantage plan.
Try it at Home
I created a Shiny app to allow users to plug in the numbers for their own plans. It is rudimentary, but it gives a sense for the relative costs of different plans.
In addition to the traditional Pi Symbol atop the cake we added Albert Einstein since today is also his birthday. It seems fitting that we lost one of the world’s other greatest physicists, Stephen Hawking on the same math holiday.
The crew has grown quite large from the five of us who celebrated our first pie day almost a decade ago.
Snowstorm Stella impacted both our numbers and our location, but last night a smaller crew braved the cold weather and messy streets to celebrate Pi Day with pizza and Pi Cake at Ribalta.
We naturally ate a lot of round pies and even a rectangular pie to honor Hippocrates’ squaring the lune.
This year’s Pi Cake came from Empire Cakes for thethirdyearinarow. It was their Brooklyn Blackout cake with Chocolate frosting, a blue Pi symbol on top and blue circles with red radii around the sides.
Last night we celebrated Rounded Pi Day by rounding at the 10,000’s digit to get 3.1416 which nicely works with the date 3/14/16. This was great after Mega Pi Day worked out so perfectly last year. And this all built uponpreviousyears’celebrations.
We ate a large quantity of pizza at Lombardi’s. and for the second year in a row we got the Pi Cake from Empire Cakes with peanut butter and chocolate flavors. The base was inscribed with historic approximations of Pi: 25/8, 256/81, 339/108, 223/71, 377/120, 3927/1250, 355/113, 62832/20000, 22/7.
A friend recently posted the following the problem:
There are 10 green balls, 20 red balls, and 25 blues balls in a a jar. I choose a ball at random. If I choose a green then I take out all the green balls, if i choose a red ball then i take out all the red balls, and if I choose, a blue ball I take out all the blue balls, What is the probability that I will choose a red ball on my second try?
The math works out fairly easily. It’s the probability of first drawing a green ball AND then drawing a red ball, OR the probability of drawing a blue ball AND then drawing a red ball.
The problem starts with three doors, one of which has a car and two of which have a goat. You choose one door at random and then the host reveals one door (not the one you chose) that holds a goat. You can then choose to stick with your door or choose the third, remaining door.
Probability theory states that people who switch win the car two-thirds of the time and those who don’t switch only win one-third of time.
But people often still do not believe they should switch based on the probability argument alone. So let’s run some simulations.
This function randomly assigns goats and cars behind three doors, chooses a door at random, reveals a goat door, then either switches doors or does not.
monty <- function(switch=TRUE)
# randomly assign goats and cars
doors <- sample(x=c("Car", "Goat", "Goat"), size=3, replace=FALSE)
# randomly choose a door
doorChoice <- sample(1:3, size=1)
# get goat doors
goatDoors <- which(doors == "Goat")
# show a door with a goat
goatDoor <- goatDoors[which(goatDoors != doorChoice)]
# if we are switching choose the other remaining door
# otherwise keep the current door
Now we simulate switching 10,000 times and not switching 10,0000 times
Continuing the annualtradition of Pi Cakes from Chrissie Cook we have gotten another Pi Cake! This year we let Drew Conway’s wife pick the flavors and she went with vanilla and red velvet (the blue color is to cause some cognitive dissonance). Looking forward to enjoying this tonight after some pizza.
With tonight’s Mega Millions jackpot estimated to be over $640 million there are long lines of people waiting to buy tickets. Of course you always hear about the probability of winning which is easy enough to calculate: Five numbers ranging from 1 through 56 are drawn (without replacement) then a sixth ball is pulled from a set of 1 through 46. That means there are choose(56, 5) * 46 = 175,711,536 possible different combinations. That is why people are constantly reminded of how unlikely they are to win.
But I want to see how likely it is that SOMEONE will win tonight. So let’s break out R and ggplot!
As of this afternoon it was reported (sorry no source) that two tickets were sold for every American. So let’s assume that each of these tickets is an independent Bernoulli trial with probability of success of 1/175,711,536.
Running 1,000 simulations we see the distribution of the number of winners in the histogram above.
So we shouldn’t be surprised if there are multiple winners tonight.
Shortly after the Giantsfantasticdefeat of the Patriots in Super Bowl XLVI (I was a little disappointed that Eli, Coughlin and the Vince Lombardi Trophy all got off the parade route early and the views of City Hall were obstructed by construction trailers, but Steve Weatherford was awesome as always) a friend asked me to settle a debate amongst some people in a Super Bowl pool.
We have 10 participants in a superbowl pool. The pool is a “pick the player who scores first” type pool. In a hat, there are 10 Giants players. Each participant picks 1 player out of the hat (in no particular order) until the hat is emptied. Then 10 Patriots players go in the hat and each participant picks again.
In the end, each of the 10 participants has 1 Giants player and 1 Patriots player. No one has any duplicate players as 10 different players from each team were selected. Pool looks as follows:
Winners = First Player to score wins half the pot. First player to score in 2nd half wins the remaining half of the pot.
The question is, what are the odds that someone wins Both the 1st and 2nd half. Remember, the picks were random.
Before anyone asks about the safety, one of the slots was for Special Teams/Defense.
There are two probabilistic ways of thinking about this. Both hinge on the fact that whoever scores first in each half is both independent and not mutually exclusive.
First, let’s look at the two halves individually. In a given half any of 20 players can score first (10 from the Giants and 10 from the Patriots) and an individual participant can win with two of those. So a participant has a 2/20 = 1/10 chance of winning a half. Thus that participant has a (1/10) * (1/10) = 1/100 chance of winning both halves. Since there are 10 participants there is an overall probability of 10 * (1/100) = 1/10 of any single participant winning both halves.
The other way is to think a little more combinatorically. There are 20 * 20 = 400 different combinations of players scoring first in each half. A participant has two players which are each valid for each half giving them four of the possible combinations leading to a 4 / 400 = 1/100 probability that a single participant will win both halves. Again, there are 10 participants giving an overall 10% chance of any one participant winning both halves.
Since both methods agreed I am pretty confidant in the results, but just in case I ran some simulations in R which you can find after the break.